If $f(x)=ax+b$ and $f^{-1}(x)=bx+a$ with $a$ and $b$ real, what is the value of $a+b$?
Answer: Since $f(f^{-1}(x))=x$, it follows that $a(bx+a)+b=x$, which implies $abx + a^2 +b = x$. This equation holds for all values of $x$ only if $ab=1$ and $a^2+b=0$.

Then $b = -a^2$.  Substituting into the equation $ab = 1$, we get $-a^3 = 1$.  Then $a = -1$, so $b = -1$, and  \[f(x)=-x-1.\]Likewise  \[f^{-1}(x)=-x-1.\]These are inverses to one another since  \[f(f^{-1}(x))=-(-x-1)-1=x+1-1=x.\]\[f^{-1}(f(x))=-(-x-1)-1=x+1-1=x.\]Therefore $a+b=\boxed{-2}$.